Browse Questions

# Integrate : $\int \limits_0^{\infty} \large\frac{1}{(x^2+g^2)^7}dx$

(a) m,n $\in$ even $k = \large\frac{\pi}{2}$

(b) m,n $\in$ even $k =2$

(c) m,n $\in$ even $k = 1$

(d) m,n $\in$ odd $k =\pi$

$x= g \tan \theta$
$dx= g \sec^2 \theta.d \theta$
$x=0, \theta_1=0$
$x=\infty, \theta_2=\large\frac{z}{2}$
$\qquad= \int \limits_0^{\large\frac{z}{2}} \large\frac{1}{(g^2)^7. (\sec^2 \theta)^7}.$$dx \qquad= \int \limits_0^{\large\frac{z}{2}} \large\frac{g}{(g^2)^7}.\frac{\sec^2 \theta. d \theta}{\sec^{14} \theta} \qquad= \int \limits_0^{\large\frac{z}{2}} \large\frac{g}{g^{14}}$$ \cos ^{13} \theta. d\theta$
$\qquad= \large\frac{g}{g^{14}} \int \limits_0^{\large\frac{z}{2}} $$\cos^{13} \theta.d \theta, n=13,k=1 => \large\frac{1}{g^{13}} \frac{(13-1)(13-3)(13-5)(13-7)(13-9)(13-11)}{13(13-2)(13-4)(13-6)(13-8)(13-10)(13-12)}.$$1$
=> $\large\frac{1}{g^{13}}. \frac{12.10.8.6.4.2}{13.11.9.7.5.3.1}$
Solving this we get,
=> $\large\frac{(8)^3.2}{3003 \times (g)^{13}}$
Hence a is the correct answer.